12C9 = 220
There are 36 possible combinations.
one digit: 5 ways two digits: 5 * 9 ways three digits: 5 * 9 * 8 ways four digits: 5 * 9 * 8 * 7 ways ... ten digits: 5 * 9! ways So if you have to use ten digits, then the answer is 5 * 9! = 1814400. If you can use anywhere from one to ten digits, then the answer is: 5 * (9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1) = 2045565
There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3
Assuming the 20 are all different, there are 20C11 = 20!/(11!*9!) = 167,960 ways.
9*8*7*6*5=15,120 ways There are 9 selections to choose from for the first song, 8 for the second, 7 for the third, 6 for the fourth, and 5 for the fifth.
(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2) ie 792 ways.
12C9 = 220
Two books are compulsory, so the student really wants to know how many ways he can pick 3 out of 7. That is 35.
There are 36 possible combinations.
72
one digit: 5 ways two digits: 5 * 9 ways three digits: 5 * 9 * 8 ways four digits: 5 * 9 * 8 * 7 ways ... ten digits: 5 * 9! ways So if you have to use ten digits, then the answer is 5 * 9! = 1814400. If you can use anywhere from one to ten digits, then the answer is: 5 * (9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1) = 2045565
There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3There are many, many ways. Here is one:9*(1+4)*(5+7+8)+6-2*3
Assuming the 20 are all different, there are 20C11 = 20!/(11!*9!) = 167,960 ways.
10 x 9 x 8 = 720.
number of ways = (7 x 6 x 5) x (9 x 8 x 7) = 105840
362880 ways. 9!=9*8*7*6*5*4*3*2*1=362880